Question

2. FAt=p (remember p=mv). A 2.5Jg baseball moving at 8.5m/s to the left is slowed by a tipped
catch to 7.5m/s to the right in 0.25 seconds, what was the force of the tipped catch?

Answer 1

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

Data obtained from the question

• Initial velocity (u) = 8.5 m/s
• Final velocity (v) = 7.5 m/s
• Time (t) = 5 ms = 0.25 s
• Mass (m) = 2.5 Kg
• Force (F) = ?

How to determine the force

The force exerted on the ball can be obtained as follow:

F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

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